Question: Let $g(x)=\dfrac{3}{x^2}+\dfrac{1}{x}$. $g'(3)=$
The strategy We can first rewrite each rational term of $g$ as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have $g'(x)$, we can plug $x=3$ into it to find $g'(3)$. Rewriting rational terms as negative powers $\begin{aligned} g(x)&=\dfrac{3}{x^2}+\dfrac{1}{x} \\\\ &=3x^{-2}+x^{-1} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}(3x^{-2}+x^{-1}) \\\\ &=3\dfrac{d}{dx}(x^{-2})+\dfrac{d}{dx}(x^{-1}) \\\\ &=3(-2x^{-3})+(-1)x^{-2} \\\\ &=-6x^{-3}-1x^{-2} \\\\ &=-\dfrac{6}{x^3}-\dfrac{1}{x^2} \end{aligned}$ Evaluating $g'(x)$ So we found that $g'(x)=-\dfrac{6}{x^3}-\dfrac{1}{x^2}$. Let's plug $x=3$ into $g'$ : $\begin{aligned} &\phantom{=}g'(3) \\\\ &=-\dfrac{6}{(3)^3}-\dfrac{1}{(3)^2} \\\\ &=-\dfrac{6}{27}-\dfrac{1}{9} \\\\ &=-\dfrac{2}{9}-\dfrac{1}{9} \\\\ &=-\dfrac39 \\\\ &=-\dfrac13 \end{aligned}$ In conclusion, $g'(3)=-\dfrac13$.